EL

btw' date=' is thermodynamics related to thermochemistry??

I am reading about this on a book called College Chemistry By Jerome L. Rosenberg and Lawrence M. Epstein.

Albert[/quote']

 

Very good; now keep reading.

Guarantee no one will solve it!

Consider two breeding strategies of the fictional Furble. Dominator Furbles can fight for a breeding territory' date=' and if they win, will be able to rear 10 offspring. An alternative is to share territory with another Furble which will allow each to rear 5 offspring. Sharers who attempt to share with dominators will be forced out of the territory, although they will be able to find a new territory. Assume sharers become extra cautious after encountering a dominator and so will always find another territory to share the next time around, but due to lost time will only be able to produce 3 offspring. Dominators are always able to force sharers out of the territory and rear 10 young. Dominators who meet dominators will win 50% of the time. When they lose, they are not able to reproduce that season due to sustained injuries. Individual Furbles cannot switch strategies.

With a total population of 2000 dominator and sharer Furbles, how many would you expect to be dominators?[/quote']

 

Let us build a legend.

Let {T} denote one season

Let {D} denote Dominator

Let {S} denote Sharer

Let {.} denote "encounters"

D.D = (50/100) * (10 + 0) = 5D/T

S.S = 5S/T

D.S = 10D/T + 3S/T = (10D + 3S)/T

The big picture here is that D are central, while S are peripheral and the population is always expanding.

From the equations above, we notice that 2/3 of the D will successfully breed, 20D/6D (assuming heterosexual breeding) because territorial behaviour is only harboured by such type of animals.

Be very careful with the terms of constant breeding, where Sharers will "always find another territory to share the next time around"; In other words New Sharers will never confront Dominators and will breed at a constant rate of 9S/6S while older Sharers are driven off and will also breed at 9S/6S consequently.

 

The colony should have two radii, r, which is the central dominance zone radius and R, which is the total population colony radius. If we do not care about calculating confrontation encounters, then we calculate for a population of 12 Furbles breeding 20D and 9S. If we do, then the peripheral D-couples will breed 10D outwards while the central D will have 50% only breeding 10D per couple.

The dominators dominate pi r^2 territorial breeding area while the sharers share ((pi R^2) - (pi r^2))territorial breeding area. R^2 must be directly proportional to the whole population, while r^2 is only directly proportional to the dominators. Notice that (2pi r) is the factor of the mixed confrontation border at which dominators expand their territory at the expense of the sharers, while (2pi R) represents the new territorial possibilities.

Having 2000 Furbles implies that 1000 Furble-couples are directly proportional to (pi R^2).

Use the rules to extract rate values for r and R per season then build an area relation with breeding ground to estimate the stable rate of growth of each type, and finally calculate the precise percentage of each.

I have a spherical ball. It is also blue. Is it more blue than it is spherical?

Let me try.

There are two critical adjectives here.

The object has a spherical shape and a blue colour.

The logical comparison operator {more} implies {less}.

It is illogical to compare colour with shape, but it is logical to compare the values of each on its own scale.

Since colour does have a scale of chromaticity then the degree of being blue is variable.

Shape, on the other hand, may not vary geometrically but may vary topologically.

This means that an egg is considered to be a deformed spherical shape that reaches its maximum perfection at zero deformity by being a perfect sphere. Spherical perfection has nothing to do with surface qualities as it is measured from centre to the mean regular surface.

Therefore a sphere is expected to be at the maximum of its topological scale when it is equated with its geometrical state.

The colour chromaticity is also defined by instrumental measuring devices, but the qualia may differ from one observer to the other.

For fairness, both shape and colour must be perceived by light reflection and by the observer's eye.

Relativistically speaking, we may get length contraction along with frequency shift deforming both qualities proportionally.

*************************************

The philosophical lesson here is the verdict.

If the observer decides that the object was a blue sphere, as it was given, then the verdict should be invariant for the session, hence the object must be equally blue and spherical or else it would have never been presented as a blue sphere from the very beginning.

The equity here is that of the invariant verdict and not of the perceptional quality of variance.

My best wishes.

*************************************

I am wondering if he had it originally as a lead accumulator electrolyte.

Perhaps his car battery is of such a type that needs sulphuric acid at such a strength.

This does not work with acid of concentration 3.75 mol/liter. Then the acid first needs to be concentrated by boiling away the water, until faint white fumes are produced. That, however, is quite a dangerous thing to do without the proper equipment (boiling the H2SO4 in a pan in the kitchen does not seem to be a good idea to me ).

[EL]

He needs a reflux column like this but in glass and with an open end.

<http://www.moonshine-still.com/page15.htm >

Then water could be boiled off.

Perhaps buying a very good book might be too demanding on a student's budget or savings.

In my country, imported hard-paperbacks in colourfully illustrated chemistry books could be as expensive as 250 dollars per issue.

Here is an excellent downloadable alternative (free of charge).

<http://www.ntu.edu.au/education/wardonli.htm >

I don't want to get into a lengthy argument based on our independent interpretations of an ill-defined question.

Now on that, we can easily agree, can't we?

What I said in my post is not what you have understood of it, so I'll be happy to clear it up, if required.

No need for that, because I did understand it quite clearly, and I appropriately responded, that the bond energy that forms between two reactants is "irrelevant" to the activity-coefficient of the reactants. Now you might wish to say that my correct response is not that to what you said. However, at an advanced level, there is an indirect relevancy between the newly formed bond energy and the _enthalpy-change-sum_ of the reactants, which was not asked, and you hinted at, but is not related to "reactivity".

In short, let it go.

Kindest regards.

he`s not trolling, he`s always asked many questions like these since I`ve known him, he`s perfectly safe

That is OK, because I gave him the benefit of the doubt and tried to answer his "question".

For some reason he was confusing between the concept of a reaction and the reactants (substrates). There are enzyme catalyzed reactions that even a bond-energy tracing would be very difficult for a beginner. That is why I was in doubt, because if you do know that you are a beginner, you usually ask more basic questions that you feel that you know how to ask them. His curiosity is very fine, yet having no faith in our patience could be his real problem, that is why he is making short cuts and jumping to conclusions upon which he is asking the wrong questions.

To Albert: Slow down, and digest each point you get an answer for before you proceed to another point. However, as {woelen} told you, there is no replacement for a small library to cover each side of your interests professionally, then if you could not understand a passage in such books come back and ask and we will gladly help you, but show some seriousness please.

5 volts are too low if you are using very diluted NaOH solution along with that narrow tube bridge away from the electrodes. The rules are:

1- Higher voltage is better.

2- Electrolyte concentration for effective electrolysis 5% to 10 % by weight (My experience)

3- The electrodes should "see" each others, which means that the solution from one electrode to the other should be uninterrupted physically by insulating material, because the solution must be linearly polarized.

I think that your device is in violation of #3 only.

Albert, there is something unsettling about your posts that gives me the impression that you are trolling.

You have somehow mastered asking the wrong questions and the experts are being kind and falling for it.

There is no such thing as "the energy level decreases in product from the reactants".

A reaction is exothermic if the ambient medium of the reaction gains heat.

A reaction is endothermic if the ambient medium of the reaction loses heat.

Now you might have believed that one should wonder, if the energy that the medium gained is not coming from the reactants as a source, then from where! We could boil a solution (adding heat) to force the reaction. This means that the product can gain heat from the flame of a Bunsen Burner. Now keeping this in mind, the product can have an enthalpy showing an increase in the energy of the product while the reactants are not even the source of the energy that was absorbed, and the reaction is endothermic. To the contrary, we might have a spontaneous combustion triggered by a nano-catalyst and it would be obviously exothermic while the starting compound loses energy to the medium. However, the energy of the product compounds could have a sum of energies that exceed the energy-difference from the combustible material as it gains back heat to trigger the spontaneous reaction.

A reaction does not necessarily have any products; surprised, right?

Dissolve some Urea in water and it will cool down the temperature of the solution, thus we have an endothermic reaction, but what are the products? Urea is still urea and water is still water but the dissolution reaction is endothermic because the hydrogen bond formation demanded some energy to form.

Now do not rush to conclusions here, because dissolving NaOH or H2SO4 in water would raise the temperature of the solution (exothermic), and here the hydrogen bond formation expels of the excess of energy.

If you wish to consider the solvent and the solute as reactants and the hydrated solution as the product, then please realise that they both define the reactants and the product and the environment as well, and here we can neither say that they gain or lose energy at all, because all is contained within the boundary.

Now we can consider some of the water as a reactant and some of the water as the environment and thus expand the boundary, but there is no way to separate the energy within the whole boundary.

So please remember that neither the reactants nor the products are exothermic or endothermic, it is the reaction that is and across the boundary that is.

DQW,

I understand what you are saying.

However, let me give you an example to show you that the relevancy is neither directly proportional nor inversely proportional as the relation between the energy of a bond and the reactants activity-coefficient at a specific temperature and concentration.

A simplest example should be HCl and NaOH.

HCl, at 25 decrees C., have the activity-coefficient varying between 0.796 at 0.1 molality to 0.809 at 1.0 molality.

NaOH, at 25 decrees C., have the activity-coefficient varying between 0.766 at 0.1 molality to 0.678 at 1.0 molality.

NaCl ionic bond and H2O hydrogen bond are (at 298 K) has a dissociation energy of 412.1 kJ/mol and 427.6 kJ/mol, respectively.

So now you have the unrelated data. What on earth can you conclude from one to give you information about the other?

The bond strength is obviously a constant, while the activity-coefficient is obviously a variant.

In other words, regardless of the conditions that causes the variance of activity (reactivity) the products have a constant bond strength per ambient condition. So what is it that you were implying to be a correlation?

I am sorry but that does not make sense either.

I could not answer him saying "irrelevant" because it is relevant, but in a different way than he is expecting.

Reactivity though has nothing to do with the final bond energy and I am not even aware of any equation that links both, so post one if you have one.

Do bond energies tell how energetic the reactants are??

Or are they irrevalent??

I am sorry to tell you that your question is incomprehensible.

There is no meaning for "How energetic reactants are?"

The reason is that the bond energy is there after reactants react.

As for "how do they react", has no standard reference to tell us how "energetic" they are.

Chemical reactions are so diverse that it would be seriously funny to even think that there is such a thing as a standard behaviour of reactants.

Sorry again for my harsh words.

OMG!

A catalyst is a substance that accelerates the rate of a chemical reaction without itself being transformed or consumed by the reaction; that is it participates in the reaction but is neither a chemical reactant nor a chemical product.

Major catalysis groups

* Converter catalysis

* Coordination catalysis

* Enzymes - biocatalysts

* Nanomaterial based surface catalysis

The major symbolic action of catalysts C.

A + C → AC (1)

B + AC → AB + C (2)

A + B + C → AB + C

The net enthalpy of C is supposed to be zero but this is not true for the rest of the reactants.

In exothermic reactions a catalyst may trigger a spontaneous reaction.

In endothermic reactions, the addition of energy from an external source is channelled through the catalyst in catalyzed reactions.

Did you read it and understood what was said?

not even noticed.

You are ignor ..... ed.

There are two more good oil-and-grease-removers not mentioned in this thread.

Orange-peal-oil and acidic-aromatic-esters.

I shall leave it as a quiz to investigate the commercial or the specific chemical compositions.

All who gave advise to use NaOH were correct, and all who said acid are fantabulously WRONG.

Acid corrodes concrete.

As per {jdurg}'s remark, he is absolutely correct, because given enough time lipids will form epoxy polymers that are very hard to strip with anything less than conc. ammonia and hydrogen peroxide used sequentially on spot or some chlorinated paint strippers produced by industry.

But what puzzles me is that the OP said "Motor Oil" which is supposed to be a saturated hydrocarbon.

If that was the case really, non of the above should have worked, while carbon black powder and liquid octane should have worked adsorption miracles.

NaOH will saponify fatty acids not parafines.

This makes me wonder if the OP really knew what the stains were caused by!

take, for example, water, and oxygen, do they react??

No.

Primarygun,

<http://www.scienceforums.net/forums/showthread.php?p=183297#post183297 >

What do you know about Supramolecular Chemistry?

But yeah what did it taste like?

 

~Scott

Gallium?

You are always welcome, latentheat.

This reminds me of the one sin sinner lighting a candle and the two sin sinner lighting two.

When Bush J. went for confession, the priest was worried about burning down the church.

He must have a very expensive "Dream Lab".